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Calculate [HY^3-] in a solution prepared by mixing 10.00 mL of 0.010 0 M VOSO₄, 9.90 mL of 0.010 0 M EDTA, and 10.0 mL of buffer with a pH of 4.00

A 25.00-mL sample containing Fe³⁺ and Cu²⁺ required 16.06 mL of 0.050 83 M EDTA for complete titration. A 50.00-mL sample of the unknown was treated with NH₄F to protect the Fe³⁺. Then Cu²⁺ was reduced and masked by thiourea. Addition of 25.00 mL of 0.050 83 M EDTA liberated Fe³⁺ from its fluoride complex to form an EDTA complex. The excess EDTA required 19.77 mL of 0.018 83 M Pb²⁺ to reach a xylenol orange end point. Find [Cu²⁺] in the unknown.

Calculate pCu²⁺ (to the 0.01 decimal place) at each of the following points in the titration of 50.0 mL of 0.040 0 M EDTA with 0.080 0 M Cu (NO₃)₂ at pH 5.00: 0.1, 5.0, 10.0, 15.0, 20.0, 24.0, 25.0, 26.0, and 30.0 mL. Make a graph of pCu²⁺ versus volume of titrant.

Calculate the concentration of H₂Y^2- at the equivalence point in Exercise 12-C

State the purpose of an auxiliary complexing agent and give an example of its use.

According to Appendix I, Cu²⁺ forms two complexes with acetate:

$\begin{array}{l}\mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{−}}\mathbf{⇌}\mathbf{C}\mathbf{u}{\left(C{H}_{3}C{O}_2\right)}^{\mathbf{+}}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}{\mathbf{β}}_{\mathbf{1}}\left(=K_1\right)\\ \mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{−}}\mathbf{⇌}\mathbf{C}\mathbf{u}{\left(C{H}_{3}C{O}_2\right)}_{\mathbf{2}}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}{\mathbf{β}}_{\mathbf{2}}\end{array}$

According to Appendix I, Cu²⁺ forms two complexes with acetate:

$\begin{array}{c}\mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{−}}\mathbf{⇌}\mathbf{C}\mathbf{u}{\left(C{H}_{3}C{O}_2\right)}^{\mathbf{+}}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}{\mathbf{β}}_{\mathbf{1}}\left(=K_1\right)\\ \mathbf{C}{\mathbf{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{−}}\mathbf{⇌}\mathbf{C}\mathbf{u}{\left(C{H}_{3}C{O}_2\right)}_{\mathbf{2}}\mathbf{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}}{\mathbf{β}}_{\mathbf{2}}\end{array}$

(a) Referring to Box 6-2, find K₂ for the reaction

$\mathit{C}\mathit{u}{\left(C{H}_{3}C{O}_2\right)}^{\mathbf{+}}\mathbf{+}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{O}}_{\mathbf{2}}^{\mathbf{−}}\mathbf{⇌}\mathit{C}\mathit{u}{\left(C{H}_{3}C{O}_2\right)}_{\mathbf{2}}\left(aq\right)\mathbf{\text{ â¶Ä‰â¶Ä‰}}{\mathit{K}}_{\mathbf{2}}$

(b) Consider 1.00 L of solution prepared by mixing 1.00 × 10^-4 mol Cu(ClO₄)₂ and 0.100 mol CH₃CO₂Na. Use Equation 12-16 to find the fraction of copper in the form Cu²⁺

Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL